There is a ring with constant changing perpendicular magnetic field producing emf = 12 V across the ring. (whole ring is an uniform resistance) If we put an ideal voltmeter connecting 1/4th of the ring then voltange showing will be?

A) 3 V (if we see the 1/4th part between voltmeter) B) 9 V (if we see the 3/4th part between voltmeter) C) 0 V (if we consider emf caused by changing field is in the form of infinite batteries and if we traverse the circut then current in infinitesimal resistance will cancel voltage due to infinitesimal battery)

I am confused please help

enter image description here

  • $\begingroup$ Where is the voltmeter? Where are the leads? How far does the field extend? $\endgroup$ – user137289 Feb 29 '20 at 14:39
  • $\begingroup$ Hi Pieter, I just added an image to make Q more clear.......field in uniformly distributed inside the ring and emf due to it is given..................................................................................After thinking for a while I think ans is 3 V becuase when we consider 3/4 th part of ring then ....V+9=12....................and when we consider 1/4th part then V+3=0 $\endgroup$ – Mitul Agrawal Feb 29 '20 at 14:59

Fools rush in where angels fear to tread...

The difficulty is that it is a non-conservative electric field that drives charge round the ring, so we cannot apply the concept of potential difference.

Nonetheless, we can calculate the current to be 12 V /R in which R is the ring resistance, and hence the voltage drops over 3/4 and 1/4 of the ring circumference to be $\frac{12\ \text V}{R} \times \tfrac34 R = 9 \text V$ and $\frac{12\ \text V}{R} \times \tfrac14 R = 3\text V$ respectively, assuming uniform resistance.

If the voltmeter is positioned as shown, it will read 3 V as there is no changing flux linking the loop that consists of bottom right quarter arc of the ring, voltmeter and voltmeter leads, and therefore no emf in this loop; all we have is the voltage drop across the quarter arc.

You object: suppose we consider the loop consisting of the other 3/4 of the ring, voltmeter and voltmeter leads? Then we have a 9 V voltage drop across the 3/4 arc, but we also have the full emf of 12 V, because the whole of the changing flux is linked with this loop. Therefore the net voltage is 3 V, as we found before.

  • $\begingroup$ If we assume the field exists only inside the ring, then there will be no flux through the loop which includes the voltmeter as long as the voltmeter stays outside of the ring (and in the plane of the ring). So, which voltage are you measuring, the 3 or the 9? (Note: The meter acts as a cut in that loop.) $\endgroup$ – R.W. Bird Mar 2 '20 at 15:58
  • $\begingroup$ Where would you measure the 12 volts? $\endgroup$ – R.W. Bird Mar 2 '20 at 16:10
  • $\begingroup$ If my line of argument is correct, with the voltmeter connected as shown, it'll show 3 V, but if placed to the left of the ring, with leads connecting it to the same places on the ring (that is at 3 o' clock and 6 o' clock) it will read 9 V. To read (nearly) 12 V, we'd keep it to the left of the ring, but move its connections to the ring closer together, say by moving the 6 o' clock connection to 16 minutes past. $\endgroup$ – Philip Wood Mar 2 '20 at 18:43

The correct answer is, C. There is no measurable voltage between any two points in this conducting ring. For each very short segment, the emf is dissipated to current flowing through resistance. Knowing the emf and R you can calculate the current and power, but find no voltage drop. If you cut the ring at some point, then the current will stop and a 12 volt drop will appear across the gap at the cut.

  • $\begingroup$ Which of us is going to try this experimentally? $\endgroup$ – Philip Wood Feb 29 '20 at 20:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.